You have found the following ages (in years) of all 4 turtles at your local zoo: $ 80,\enspace 55,\enspace 55,\enspace 54$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{80 + 55 + 55 + 54}{{4}} = {61\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $80$ years $19$ years $361$ years $^2$ $55$ years $-6$ years $36$ years $^2$ $55$ years $-6$ years $36$ years $^2$ $54$ years $-7$ years $49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{361} + {36} + {36} + {49}} {{4}} $ $ {\sigma^2} = \dfrac{{482}}{{4}} = {120.5\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{120.5\text{ years}^2}} = {11\text{ years}} $ The average turtle at the zoo is 61 years old. There is a standard deviation of 11 years.